# Why is $\mathcal{O}_K$ the ring to be considered for factorizing integers?

For $K = \mathbb{Q}[\alpha]$ (with $\alpha$ algebraic over $\mathbb{Q}$), I understand that $\mathbb{Z}[\alpha]$ may be too coarse, and that $\mathcal{O}_K$ (the algebraic integers of $K$) allows more accurate factorizations into irreductible (not necessarily prime) factors.

But I do not understand why $\mathcal{O}_K$ is the finest subring of $K$ to be considered. Why is the definition of an algebraic integer: it has a monic minimal polynomial in $\mathbb{Z}[X]$ ? Why are no more algebraic numbers of $K$ interesting for factorizing ? Why should the set of the numbers considered be a ring (since only the multiplication matters) ?

For instance, $7 = \frac{5 + \sqrt{-3}}2 \cdot \frac{(5 - \sqrt{-3})}2$ is a "proper" factorization, while $7 = \frac{7}2 \cdot 2$ is "improper". How can a "proper" factorization be characterized ? How does this characterization define $\mathcal{O}_K$ ? I am thus looking of an alternative (and most likely equivalent) definition of $\mathcal{O}_K$, based on the idea of "proper" factorization.

Related questions: How essential is the fact that the integers of $K$ are a finitely generated $\mathbb{Z}$-module ? If $f$, a monic polynomial in $\mathbb{Z}$[X], has a monic factor in $K[X]$, does this factor already belong to $\mathcal{O}_K[X]$ ?

• Doesn't $\frac{7}{2} \times 2$ strike you as somewhat redundant and inelegant? Although I suppose you could also say that about $(\sqrt 7)^2$. – Mr. Brooks Aug 29 '17 at 22:09

I'm not sure I can give an exhaustive answer, but I will try and explain the things that occurred to me while reading your question.

1. Why we consider $\mathcal{O}_K$ rather than $\Bbb{Z}[\alpha]$? This is partly because $\mathcal{O}_K$ is "well defined". There are many algebraic integers $\beta\in K$ such that $K=\Bbb{Q}(\beta)$. So which of those $\beta$s should we use? They may (and will!) lead to different rings $\Bbb{Z}[\beta]$ meaning that the resulting ring would not be an invariant of the field $K$. When we go up to $\Bbb{O}_K$ we take the union of all those rings $\Bbb{Z}[\beta]$! This is surely more democratic! Also better behaved in the sense that the result only depends on the field $K$ rather than on the way we happened to construct it!
2. Why we want to look at only rings? The reason for this was not immediately clear earlier in the history, and is actually rather delicate. When looking at factorizations we have problems like the following. Consider $K=\Bbb{Q}(\sqrt{-5})$ when $\mathcal{O}_K=\Bbb{Z}[\sqrt{-5}]$. In this ring have factorizations like $2\cdot3=6=(1+\sqrt{-5})(1-\sqrt{-5})$. Furthermore, none of $2,3,1\pm\sqrt{-5}$ split further into smaller factors. To deal with such non-uniqueness of factorizations Kummer first introduced so called ideal numbers. That was clumsy to work with, and was quickly replace by the concept of ideals (check that WP page for more about the history). It turned out that using ideals we get a nicer theory of factorization. But, ideals need an ambient ring!
3. Why not consider bigger subrings of $K$? We can, and sometimes we do! But when we do that many primes drop out of the picture. This shows already at the level of $\Bbb{Z}$. If instead of that subring of $\Bbb{Q}$ we use a bigger ring like $$\Bbb{Z}[1/2] = \left\{\frac{m}{2^k}\mid m,k\in\Bbb{Z}\right\},$$ what happens is that in a sense $2$ stops being a prime. It becomes a unit instead. Therefore the primes of the ring $\Bbb{Z}[1/2]$ are $3,5,7,\cdots$. A common way of using this kind of rings is to isolate the study to a so called local case, when we have only a single prime remaining. An example of such a ring would be $$\Bbb{Z}_{(2)} = \left\{\frac{m}{n}\mid m, n\in\Bbb{Z}, \text{n odd}\right\}.$$ In this ring $2$ is the only remaining prime. If $p>2$ is a prime number, then $1/p$ is an element of this ring, so it has become a unit! Similar thing is often done to construct maximal subrings of $K$.
4. Why not allow factorizations like $7=\dfrac72\cdot2$? If we do that inside a ring, then $\dfrac12=\dfrac72-3$ must also be included. This means that we are actually working with the ring $\Bbb{Z}[1/2]$. Meaning that $2$ ceases to be a prime. So, as above, we can do this, but then we drop $2$ out of our list of primes.

None of this is probably very compelling. The upshot is that doing it this way gave us a nice theory. The concepts have evolved, and anything useful I have said has the benefit of a 20/20 hindsight.

• Thank you, your considerations are interesting. Can you please develop: "Furthermore, none of $2,3,1±\sqrt{−5}$ split further into smaller factors." That is precisely my question, why are they irreducible ? What is the criterion ? Other remark, about point 4: You assume that the elements considered are a ring, it is convenient, but why should it be necessary (only the multiplication matters) ? – Loic Aug 29 '17 at 19:05
• Inspired by your point 4: $\mathcal{O}_K$ may be the biggest superset of $\mathbb{Z}$ which is closed under multiplication, and where no prime of $\mathbb{Z}$ becomes a unit. Then it happens that it is also closed under addition. That would be an alternative definition I am looking for. And do you think that my very last question is true: A polynomial factorization in $K[X]$ is essentially already in $\mathcal{O}_K[X]$ ? It would be another good reason to consider $\mathcal{O}_K$. It seems similar to Gauss' Lemma, I will study it more deeply. – Loic Aug 29 '17 at 19:36
• @Loic As a $\mathbb{Q}$-vector space $K = \sum_{i=1}^n u_i \mathbb{Q}$ for some $u_i \in K$. Then $\mathcal{O}_K$ is the largest subring of $K$ which is also a lattice : of the form $\sum_{i=1}^n v_i \mathbb{Z}$. @ Jyrki From this, I'd like a short proof of $\prod_{\sigma \in G} \sigma(I) = (N(I))$ (if $K/\mathbb{Q}$ is Galois) allowing to discuss how (fractional) ideals are inversible. – reuns Aug 29 '17 at 19:40

First off, notion of factorization we want is not that of factoring elements of the ring, but ideals.

A relevant abstract notion is that of a Dedekind domain — a ring where every* nonzero ideal factors (uniquely!) into a product of prime ideals.

If $K$ is a number field, its ring of integers is the smallest subring with the properties that:

• $\mathcal{O}_K$ is a Dedekind domain
• $K$ is the fraction field of $\mathcal{O}_K$

If you study all subrings of $K$, you'll find that the description of the ideals of such rings all relate to the ideals of $\mathcal{O}_K$:

• If you included things that aren't algebraic integers, you'll see that the the resulting ring is missing the ideals that correspond to factors of the denominator
• If you don't include all of the algebraic integers, you'll find that some of the ideals are defective, and passing to the integral closure (which will include all of the algebraic integers) resolves those defects.

There is actually a field-theoretic description as well. In the rational numbers, the exponents of prime factorizations relative to the integers are examples of discrete valuations on $\mathbb{Q}$.

In fact, every discrete valuation corresponds to a prime of $\mathbb{Z}$, and $\mathbb{Z}$ is characterized as the subring of elements $x \in \mathbb{Q}$ with the property that $v(x) \geq 0$ for every discrete valuation $v$.

The same is true in number fields; $\mathcal{O}_K$ is precisely the of elements $x \in K$ with $v(x) \geq 0$ for every discrete valuation. And every discrete valuation corresponds to a prime ideal of $\mathcal{O}_K$.

*: The unit ideal is the empty product

• Can you please develop: "If you included things that aren't algebraic integers, you'll see that the the resulting ring is missing the ideals that correspond to factors of the denominator"; what denominator ? some example ? Furthermore, why it is illegitimate to ask about factorization in irreducible elements ? Dedekind introduced ideals for achieving unique factorization, but before, a notion of irreducibility is needed; this is the notion I want to understand better. In the classical example $6 = 2 * 3 = (1 + \sqrt{-5}) * (1 - \sqrt{5})$, why are the factors irreducible? – Loic Aug 29 '17 at 18:58

In mathematics, you can define anything at all, including things that don't actually exist. Regardless of whether the existence of what you've defined can be proven or not, it could be open to debate whether your definition is useful or at least interesting.

In your example of the factorization of 7 in $\mathbb Z[\omega]$, where $$\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2},$$ how would we define a ring "larger" than $\mathbb Z[\omega]$ such that twice seven halves is a valid factorization of 7 into irreducibles?

Clearly $\mathbb Q(\sqrt{-3})$ won't do, because then $$7 = 2 \times \frac{7}{2} = 2^2 \times \frac{7}{4} = 2^3 \times \frac{7}{8} = 2^4 \times \frac{7}{16} = \ldots$$ Actually, 2 can also be broken down further in $\mathbb Q(\sqrt{-3})$, but this is enough to make the point that $\mathbb Q(\sqrt{-3})$ is not the ring we're looking for.

In such a ring, the numbers 2, seven halves, $$\frac{5}{2} - \frac{\sqrt{-3}}{2}, \frac{5}{2} + \frac{\sqrt{-3}}{2}$$ might become irreducible but not prime, and that's okay. We should still have the expectation that if we divide any of these numbers by a unit (such as $\omega$ would still be) the result would be a number that is also in the ring.

And so we have $$\frac{2}{\omega} = -2 \omega,$$ $$\frac{\frac{5}{2} - \frac{\sqrt{-3}}{2}}{\omega} = -2 - \sqrt{-3},$$ $$\frac{\frac{5}{2} + \frac{\sqrt{-3}}{2}}{\omega} = -\frac{1}{2} - \frac{3 \sqrt{-3}}{2}$$ and $$\frac{\frac{7}{2}}{\omega} = -\frac{7}{4} - \frac{\sqrt{-3}}{4}.$$ Purely for my own convenience, I label the last of these four numbers $\theta$.

The first three of these pose no problem, since they are also in $\mathbb Z[\omega]$, but $\theta$ is clearly not. Maybe there is a way to define a ring such that $\theta$ belongs in it and the process of factorization into irreducibles can eventually reach a conclusion.

But I'm beginning to lose interest, and starting to feel adrift in the complex plane. Solving the equation $4x^2 + 14x + 49$ just doesn't seem that interesting to me now. Solving $x^2 - 5x + 7$ does interest me, though there are plenty of people who wouldn't care about the solution to that one either.